Uniqueness of Tetration - Revision history

Bo198214: work in progress

2017-01-04T08:04:23Z

work in progress

Bo198214: /* Proposition */

2017-01-03T19:44:58Z

Proposition

Bo198214 at 19:44, 3 January 2017

2017-01-03T19:44:25Z

Bo198214: /* Proof. */

2017-01-03T18:55:13Z

Proof.

Bo198214: /* Proposition */

2017-01-03T16:42:42Z

Proposition

Bo198214: /* Proof. */

2017-01-03T16:25:12Z

Proof.

Bo198214: Created page with "== Proposition == For each $b>e^{1/e}$ there exists exactly one holomorphic function $\sigma_b$ on $D=\C\setminus\{x \le -2\}$ which satisfies for all $z\in D$: $$\overline{\s..."

2017-01-03T16:20:22Z

Created page with "== Proposition == For each $b>e^{1/e}$ there exists exactly one holomorphic function $\sigma_b$ on $D=\C\setminus\{x \le -2\}$ which satisfies for all $z\in D$: $$\overline{\s..."

New page

== Proposition ==
For each $b>e^{1/e}$ there exists exactly one holomorphic function $\sigma_b$ on $D=\C\setminus\{x \le -2\}$ which satisfies for all $z\in D$:
$$\overline{\sigma_b(z)}=\sigma_b(\overline{z})$$, $$\sigma_b(z+1)=b^{\sigma_b(z)}$$ and $$\sup_{t\to\infty}\left| \sigma_b(s+{\rm i}t)\right| <\infty$$ for each $s>-2$.

==== Proof. ====
We know there exists already a solution $\tau_b$ which satisfies all conditions and its inverse $\alpha_b$ which is a whole function and satisifies $\alpha_b(b^z)=\alpha_b(z)+1$.
Then we know that the function $\delta(z)=\alpha_b(\sigma_b(z))-z$, holomorphic on $D$, is periodic with period 1:
$$\delta(z+1)=(\alpha_b(\sigma_b(z+1))-(z+1)=\alpha_b(\sigma_b(z))-z=\delta_b(z)$$
and is real for $z > -2$ (and can be continued to $\R$) and can to developed into a real Fourier-Series ($A_k$, $\phi_k$ in $\R$):

$$\delta(t)=\sum_{k=0}^{\infty}A_{k}\cos\left(2\pi kt-\varphi_{k}\right)$$

$$\delta(z)=\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(\exp\left(2\pi {\rm i}kz-{\rm i}\varphi_{k}\right)+\exp\left(-2\pi {\rm i}kz+{\rm i}\varphi_{k}\right)\right)$$

$$
\delta(s+{\rm i}t)=\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(\exp\left(2\pi {\rm i}ks-2\pi kt -{\rm i}\varphi_{k}\right)+\exp\left(-2\pi {\rm i}ks+2\pi kt+{\rm i}\varphi_{k}\right)\right)
$$

← Older revision		Revision as of 08:04, 4 January 2017
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−	This is work in progress, dont rely on it!	+	'''This is work in progress, dont rely on it!'''

	== Proposition ==		== Proposition ==

← Older revision		Revision as of 19:44, 3 January 2017
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		+	This is work in progress, dont rely on it!
		+
	== Proposition ==		== Proposition ==
	For each $b>e^{1/e}$ there exists exactly one holomorphic function $\sigma_b$ on $D=\C\setminus\{x \le -2\}$ which is real for all $x>-2$ and satisfies for all $z\in D$:		For each $b>e^{1/e}$ there exists exactly one holomorphic function $\sigma_b$ on $D=\C\setminus\{x \le -2\}$ which is real for all $x>-2$ and satisfies for all $z\in D$:

@@ Line 17: / Line 17: @@
 =\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(\cos(2\pi kx-\varphi_{k})\left(\exp\left(-2\pi k y\right)+\exp(2\pi ky)\right)+{\rm i}\sin(2\pi kx-\varphi_{k})\left(-\exp\left(2\pi k y\right)+\exp(-2\pi ky)\right)\right)$$
-$$\Im(\delta(x+{\rm i}y))=\sum_{k=0}^{\infty}\frac{A_{k}}{2}\sin(2\pi kx-\varphi_{k})\left(-\exp\left(2\pi k y\right)+\exp(-2\pi ky)\right)$$
+$$\Im(\delta(x+{\rm i}y)+x+{\rm i}y)=y+\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(-\exp\left(2\pi k y\right)+\exp(-2\pi ky)\right)\sin(2\pi kx-\varphi_{k})$$
 $$

← Older revision		Revision as of 16:25, 3 January 2017
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	\delta(s+{\rm i}t)=\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(\exp\left(2\pi {\rm i}ks-2\pi kt -{\rm i}\varphi_{k}\right)+\exp\left(-2\pi {\rm i}ks+2\pi kt+{\rm i}\varphi_{k}\right)\right)		\delta(s+{\rm i}t)=\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(\exp\left(2\pi {\rm i}ks-2\pi kt -{\rm i}\varphi_{k}\right)+\exp\left(-2\pi {\rm i}ks+2\pi kt+{\rm i}\varphi_{k}\right)\right)
	$$		$$
		+
		+	But this expression can only be bounded with respect to $t$ if $A_k=0$ for all $k\ge 1$. Hence $\delta$ is a constant $c$ and $\sigma_b(z)=\tau_b(z+c)$.

@@ Line 5: / Line 5: @@
 ==== Proof. ====
 We know there exists already a solution $\tau_b$ which satisfies all conditions and its inverse $\alpha_b$ which is a whole function and satisifies $\alpha_b(b^z)=\alpha_b(z)+1$.
-Then we know that the function $\delta(z)=\alpha_b(\sigma_b(z))-z$, holomorphic on $D$, is periodic with period 1:
+Then we know that
 $$\delta(z+1)=(\alpha_b(\sigma_b(z+1))-(z+1)=\alpha_b(\sigma_b(z))-z=\delta_b(z)$$
 and is real for $z > -2$ (and can be continued to $\R$) and can so be developed into a real Fourier-Series ($A_k$,  $\phi_k$ in $\R$):
@@ Line 12: / Line 14: @@
 $$\delta(z)=\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(\exp\left(2\pi {\rm i}kz-{\rm i}\varphi_{k}\right)+\exp\left(-2\pi {\rm i}kz+{\rm i}\varphi_{k}\right)\right)$$
 $$
-\delta({\rm i}t)=\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(\exp\left(-2\pi kt -{\rm i}\varphi_{k}\right)+\exp\left(2\pi kt+{\rm i}\varphi_{k}\right)\right)
+\alpha_b(\sigma_b({\rm i}t))=\delta({\rm i}t)+{\rm i}t={\rm i}t+\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(\exp\left(-2\pi kt -{\rm i}\varphi_{k}\right)+\exp\left(2\pi kt+{\rm i}\varphi_{k}\right)\right)
 $$
 But this expression can only be bounded with respect to $t$ if $A_k=0$ for all $k\ge 1$. Hence $\delta$ is a constant $c$ and $\sigma_b(z)=\tau_b(z+c)$.

@@ Line 1: / Line 1: @@
 == Proposition ==
-For each $b>e^{1/e}$ there exists exactly one holomorphic function $\sigma_b$ on $D=\C\setminus\{x \le -2\}$ which satisfies for all $z\in D$:
+For each $b>e^{1/e}$ there exists exactly one holomorphic function $\sigma_b$ on $D=\C\setminus\{x \le -2\}$ which is real for all $x>-2$ and satisfies for all $z\in D$:
-$$\overline{\sigma_b(z)}=\sigma_b(\overline{z})$$, $$\sigma_b(z+1)=b^{\sigma_b(z)}$$ and $$\sup_{t\to\infty}\left| \sigma_b(s+{\rm i}t)\right| <\infty$$ for each $s>-2$.
+$$\sigma_b(0)=1, \sigma_b(z+1)=b^{\sigma_b(z)}$$ and $$\sup_{t\to\infty}\left| \sigma_b({\rm i}t)\right| <\infty$$.
 ==== Proof. ====
@@ Line 7: / Line 7: @@
 Then we know that the function $\delta(z)=\alpha_b(\sigma_b(z))-z$, holomorphic on $D$, is periodic with period 1:
 $$\delta(z+1)=(\alpha_b(\sigma_b(z+1))-(z+1)=\alpha_b(\sigma_b(z))-z=\delta_b(z)$$
-and is real for $z > -2$ (and can be continued to $\R$) and can to developed into a real Fourier-Series ($A_k$,  $\phi_k$ in $\R$):
+and is real for $z > -2$ (and can be continued to $\R$) and can so be developed into a real Fourier-Series ($A_k$,  $\phi_k$ in $\R$):
 $$\delta(t)=\sum_{k=0}^{\infty}A_{k}\cos\left(2\pi kt-\varphi_{k}\right)$$
@@ Line 14: / Line 14: @@
 $$
-\delta(s+{\rm i}t)=\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(\exp\left(2\pi {\rm i}ks-2\pi kt -{\rm i}\varphi_{k}\right)+\exp\left(-2\pi {\rm i}ks+2\pi kt+{\rm i}\varphi_{k}\right)\right)
+\delta({\rm i}t)=\sum_{k=0}^{\infty}\frac{A_{k}}{2}\left(\exp\left(-2\pi kt -{\rm i}\varphi_{k}\right)+\exp\left(2\pi kt+{\rm i}\varphi_{k}\right)\right)
 $$
 But this expression can only be bounded with respect to $t$ if $A_k=0$ for all $k\ge 1$. Hence $\delta$ is a constant $c$ and $\sigma_b(z)=\tau_b(z+c)$.